Hello

I made a 2 tone audio generator with variable output, it works well. I wanted to measure the output of the audio generator so I connected it to a oscilloscope I got a 2 volt peak to peak reading. I then connected it to my digital meter and the voltage was different?? I assumed the correct voltage would be 1 volt with 2 volts peak to peak on the oscilloscope??
All this brought up the big question do I use the AC mode or the DC mode on the digital meter or the oscilloscope for that matter?

Sometimes You think you know but you don't

Thanks

 

Does your DMM do true RMS measurement? I'd guess 1.41V for 2V P to P. √2 and all that.

 

I assumed the correct voltage would be 1 volt with 2 volts peak to peak on the oscilloscope??

The meter is calibrated to measure RMS voltage which is 0.707 times the peak voltage of a sinewave.
Why did you think it was 1/2?

You use the AC mode on the multimeter and you can use either mode on the oscilloscope.
AC mode on the oscilloscope blocks any DC in the signal.

 

In a AC sine wave from zero to peak is the voltage, so peak to peak would be double ??
The meter is calibrated for RMS voltage (I just checked GRIN)

 

You set the DMM for AC.

2V peak-to-peak is 1V amplitude which is 0.7V RMS.

Then there is the matter of frequency. DMM RMS readout is only valid over a certain frequency range.
What is the frequency of your sine wave?

 

1250 Hz and 2000 Hz ( for 2 tone modulation test)

 

2V peak-to-peak is 1V amplitude which is 0.7V RMS

So 1 volt amplitude would display 0.7 volts on my dvm??
I am supposed to inject a 50mv signal for testing what would be the best way to determine that voltage??

I am sure learning a lot <GRIN>

 

50 mv what? p-p? RMS? peak?
Use resistors and the 'scope after you figure out what label the voltage is supposed to have.

 

Audio signals are sometimes referenced to 1mW into 600Ω = 0dB.

https://en.wikipedia.org/wiki/Line_level

 

So 1 volt amplitude would display 0.7 volts on my dvm??
I am supposed to inject a 50mv signal for testing what would be the best way to determine that voltage??

I am sure learning a lot <GRIN>

What make and model DMM do you have.

 

The voltage output is adjustable from 0- about 3 volts, The concern is do I use the DMM or the Oscilloscope to determine the voltage? which one is correct??

What make and model DMM do you have.

GW GDM 8135

 

Firstly, calibrate the voltage scale of the oscilloscope using a reliable DC reference.
For example, measure the voltage on a battery using the DMM on DC Volts. Compare with the vertical deflection on the oscilloscope.

Next, use the oscilloscope to measure the peak-to-peak voltage of a sine wave. Multiply the peak-to-peak voltage by 0.35.

Now compare that with the reading you get on the DMM set to AC volts.

 

Next, use the oscilloscope to measure the peak-to-peak voltage of a sine wave. Multiply the peak-to-peak voltage by 0.35.

Now compare that with the reading you get on the DMM set to AC volts.

OK it worked!!
So just to make sure I understand what I just learned, The peak amplitude on the oscilloscope show the maximum voltage at that moment in time, but prior and after that moment the voltage is increasing and decreasing. Were as DC is constant voltage. To make the AC voltage measurement equal to the dc voltage measurement we must multiply amplitude x 70%. Visually speaking pushing peak voltage down and squaring the sine- wave.
Please tell me I am right <GRIN> cause it's starting to make sense

 

OK it worked!!
So just to make sure I understand what I just learned, The peak amplitude on the oscilloscope show the maximum voltage at that moment in time, but prior and after that moment the voltage is increasing and decreasing. Were as DC is constant voltage. To make the AC voltage measurement equal to the dc voltage measurement we must multiply amplitude x 70%. Visually speaking pushing peak voltage down and squaring the sine- wave.
Please tell me I am right <GRIN> cause it's starting to make sense

Almost. Multiply the peak-to-peak sine wave voltage by 0.35 instead.

If you want to Multiply by 0.7 volts, then you have to use zero-to-peak voltage from your scope. This is not easy to find the zero point so peak-to-peak is typically used.

 

Hello

I made a 2 tone audio generator with variable output, it works well. I wanted to measure the output of the audio generator so I connected it to a oscilloscope I got a 2 volt peak to peak reading. I then connected it to my digital meter and the voltage was different?? I assumed the correct voltage would be 1 volt with 2 volts peak to peak on the oscilloscope??
All this brought up the big question do I use the AC mode or the DC mode on the digital meter or the oscilloscope for that matter?

Sometimes You think you know but you don't

Thanks

Hi: I read a little further down in the thread. Use your DMM to measure the input for testing. The generator level should be set to 50mv when connected to the input of the amp. Then measure the output level. The amplifier gain then (In DB) would be 20 times the log of the ratio of the output to the input. For example, suppose you measure 1.5 volts RMS when the input is set to 50 mv. The gain in DB would be 1.5/.05 = 30 then take the log to the base 10 of 30 = 1.477 multiplied by 20 = 29 DB. Hope this helps.

 

Thank You everyone for all the help, was a real AHH-HA moment for me

 

OK it worked!!
So just to make sure I understand what I just learned, The peak amplitude on the oscilloscope show the maximum voltage at that moment in time, but prior and after that moment the voltage is increasing and decreasing. Were as DC is constant voltage. To make the AC voltage measurement equal to the dc voltage measurement we must multiply amplitude x 70%. Visually speaking pushing peak voltage down and squaring the sine- wave.
Please tell me I am right <GRIN> cause it's starting to make sense

I see that you are now learning the meaning of RMS, root mean square.

The RMS value of any waveform is the equivalent constant DC voltage that has the same heating effect as the random waveform. Heating effect means that we are comparing power in the two signals. That is, the waveform squared and averaged over time is the same as the DC voltage squared.

How is RMS calculated? Just as the words state - root mean square.

Square - we take the square of the signal
Mean - we take the average value. If the waveform is periodic, we can take the average value over one cycle.
Root - we take the square root of the average

When we do the math for a sine wave of amplitude A, it works out to be A divided by √2.
This applies for a sine wave. For other waveforms the conversion is different.

Note that RMS value and average value are not the same for a sine wave.